Consider the equilibrium: 4 A + 2 B ⇌ 2 C + 2 D (Kc = 18.8). Calculate Kc' for the equilibrium C + D ⇌ 2 A + B.
we have given,
4A + 2B ⇋ 2C + 2D
Kc = 18.8
\(\Rightarrow\) Kc = \(\frac{[C]^2[D]^2}{[A]^4[B]^2}\) ---- (1)
C + D ⇋ 2A + B.
\(K'_c\) = \(\frac{[A]^2[B]}{[C][D]}\) ---- (2)
\(\Rightarrow\) \(\frac{1}{K'_c}=\frac{[C][D]}{[A]^2[B]}\)
\(\Rightarrow\) \(\frac{1}{(K'_c)^2}=\frac{[C]^2[D]^2}{[A]^4[B]^2}\) ------ (3)
From equation -- (1) and (3) --
Kc = \(\frac{1}{(K'_c)^2}\)
putting the value of kc = 18.8
\(\Rightarrow\) \(K'_c=\frac{1}{\sqrt{18.8}}\)
\(K'_c\) = 0.23